Trig question
- Thread starter Mstoothlady2012
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I think it is 152 degree in 2nd quadrant and it will be similar to 22 degree,0.4.
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this is weird, but i changed it to degrees first which i got 157.5 which is in the second quadrant making it a negative...its close to 150 degrees which would be -1/2 but i dont know...what are the answer choices???
a) sin 3pi/4
b) sin pi/8
c) sin pi/4
d) cos pi/8
e) cos 7pi/8
Sometimes kaplan sucks in explaining the solution 🙄
b) sin pi/8
c) sin pi/4
d) cos pi/8
e) cos 7pi/8
Sometimes kaplan sucks in explaining the solution 🙄
sin(7pi/8) is in the 2nd quadrant so it's positive. You can eliminate only (e) with this.
You need to remember your reference angles. These are (pi +- value), (2pi - value), and just (value). So if your value (angle) = pi/3, then your reference angles are (pi + pi/3 = 4pi/3), (pi - pi/3 = 2pi/3), (2pi - pi/3 = 5pi/3), and just (pi/3).
Here you have sin(7pi/8) which is the reference angle when you use pi/8 for your value, plugged into (pi - value).
Your reference angles are pi/8, 7pi/8, 9pi/8, and 15pi/8. The only one that has the same SIGN is pi/8. So you want sin(pi/8).
The other 2 reference angles (9pi/8 and 15pi/8) give you the same MAGNITUDE but the opposite sign.
Thanks streetwolf!! 🙂
great explanation just i have 1 question, how do you know which has the same sign as 7pi/8?? i know that 7pi/8 is in the second quadrant and that will give you a positive number but how do you know that the other ones are negative or positive...the hard way is to change the values to degrees and check but how do you do it without changing it to degrees...thanks
wait nvm i just figured it out haha stupid question...
pi/8 is the reference meaning its in the 1st quadrant, then 7pi/8 has to be pi - the reference angle which is 180 minus whatever it is so that has to be in the 2nd quadrant, 9pi/8 has to be the reference angle plus pi so it has to be in the 3rd quadrant which would give you a negative sin...
pi/8 is the reference meaning its in the 1st quadrant, then 7pi/8 has to be pi - the reference angle which is 180 minus whatever it is so that has to be in the 2nd quadrant, 9pi/8 has to be the reference angle plus pi so it has to be in the 3rd quadrant which would give you a negative sin...
Any angle that is within the pi range will give you positive sign.
but doesnt the reference angle of pi/8 always start in the first quadrant???
like if you have pi + pi/8 = 9pi/8 (that just means pi= 180 and your adding to it the reference angle so it would be in the 3rd quadrant) when you do pi - pi/8 its basically 180 minus that reference angle so its in the 2nd quadrant...
like if you have pi + pi/8 = 9pi/8 (that just means pi= 180 and your adding to it the reference angle so it would be in the 3rd quadrant) when you do pi - pi/8 its basically 180 minus that reference angle so its in the 2nd quadrant...
yup
The 4 reference angles are always in different quadrants. For cosine you get positive values at x > 0 which means 1st and 4th quadrants. For sine you get positive values at y > 0 which means 1st and 2nd quadrants.
You guys have made this so complicated. There is no need to convert to degrees. You should simply know that sin (pi - x) = sin x . Therefore, sin (7pi/8) = sin (pi - pi/8) = sin pi/8 .
important formulas to know:
sin (pi - x) = sin x
cos (pi - x) = - cos x
sin (pi + x) = - sin x
cos (pi + x) = - cos x
sin (Pi/2 - x) = cos x
cos (pi/2 - x) = sin x
sin (pi/2 + x) = cos x
cos (pi/2 + x) = - sin x
sin (-x) = - sin x
cos (-x) = cos x
Do not memorize these. Just draw the trignometric circle [with sin and cos axes] and try to understand these fomula. Then you can make them up during the test.
important formulas to know:
sin (pi - x) = sin x
cos (pi - x) = - cos x
sin (pi + x) = - sin x
cos (pi + x) = - cos x
sin (Pi/2 - x) = cos x
cos (pi/2 - x) = sin x
sin (pi/2 + x) = cos x
cos (pi/2 + x) = - sin x
sin (-x) = - sin x
cos (-x) = cos x
Do not memorize these. Just draw the trignometric circle [with sin and cos axes] and try to understand these fomula. Then you can make them up during the test.
